We have the following problem:
$$ u_{tt} - u_{xx} = \cos(x+t), $$
$$u(x,0) = x, $$
$$u_t(x,0) = \sin x. $$ I was told by my professor to let $u(x,t) = w(x,t) + v(x,t)$, where $v(x,t)$ is a particular solution for the above problem and $w(x,t)$ is the general solution of the homogeneous equation.
I really can't find the particular solution ($v(x,t)$), can anyone give me a way for finding a particular solution?
Note: this is not a homework, but a practice for an exam.
Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,
Then $u_x=u_pp_x+u_qq_x=u_p+u_q$
$u_{xx}=(u_p+u_q)_x=(u_p+u_q)_pp_x+(u_p+u_q)_qq_x=u_{pp}+u_{pq}+u_{pq}+u_{qq}=u_{pp}+2u_{pq}+u_{qq}$
$u_t=u_pp_t+u_qq_t=u_p-u_q$
$u_{tt}=(u_p-u_q)_t=(u_p-u_q)_pp_t+(u_p-u_q)_qq_t=u_{pp}-u_{pq}-u_{pq}+u_{qq}=u_{pp}-2u_{pq}+u_{qq}$
$\therefore u_{pp}-2u_{pq}+u_{qq}-(u_{pp}+2u_{pq}+u_{qq})=\cos p$
$-4u_{pq}=\cos p$
$u_{pq}=-\dfrac{\cos p}{4}$
$u(p,q)=f(p)+g(q)-\dfrac{q\sin p}{4}$
$u(x,t)=f(x+t)+g(x-t)-\dfrac{(x-t)\sin(x+t)}{4}$
$u_t(x,t)=f_x(x+t)-g_x(x-t)-\dfrac{(x-t)\cos(x+t)-\sin(x+t)}{4}$
$u(x,0)=x$ , $u_t(x,0)=\sin x$ :
$\begin{cases}f(x)+g(x)-\dfrac{x\sin x}{4}=x\\f_x(x)-g_x(x)-\dfrac{x\cos x-\sin x}{4}=\sin x\end{cases}$
$\begin{cases}f(x)+g(x)=x+\dfrac{x\sin x}{4}\\f_x(x)-g_x(x)=\dfrac{x\cos x+3\sin x}{4}\end{cases}$
$\begin{cases}f(x)+g(x)=x+\dfrac{x\sin x}{4}\\f(x)-g(x)=\dfrac{x\sin x-2\cos x}{4}+c\end{cases}$
$\begin{cases}f(x)=\dfrac{2x+x\sin x-\cos x}{4}+\dfrac{c}{2}\\g(x)=\dfrac{2x+\cos x}{4}-\dfrac{c}{2}\end{cases}$
$\therefore u(x,t)=\dfrac{2(x+t)+(x+t)\sin(x+t)-\cos(x+t)}{4}+\dfrac{c}{2}+\dfrac{2(x-t)+\cos(x-t)}{4}-\dfrac{c}{2}-\dfrac{(x-t)\sin(x+t)}{4}=x+\dfrac{t\sin(x+t)+\sin x\sin t}{2}$